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Potential Gradient [Question(s) regarding the concept]

  1. Sep 14, 2015 #1
    [Note from mentor: This was originally posted in a non-homework forum so it doesn't have the homework template.]

    -----------------------------------------

    Problem:
    The surface of a solid metal sphere (radius r = 4.58 cm) is at potential V = 9,851 Volts. Find the magnitude of the potential gradient at distance 9.21r from the center of the sphere (outside the sphere), in V/m.
    Assume: V(∞) = 0.


    My point: I have this general understanding that the Potential Gradient equals to ## \frac { \Delta V } { \Delta x } ## where - ## x ## - is the distance between the charge ## Q ## and the point (or another charge.)

    My question: Well in this problem, I am given ## V ## being 9851 Volts but that's on one side.
    I watched an educational (physics) video saying that if there is only one charge and no other charges, then the Electric Potential Energy equals to ## zero. ## (which makes the Potential (at distance 9.21 * r = 0 V) That's being said, should I just divide the given ## V ## by the ## distance - x - ## which is also given. [ ## 9.21 * r ## ]?

    Please help me with the concept. If you have a link to an online file that could explain this topic in depth please share it. I really want to learn but what I learned from my textbook is not leading me to find answers to problems I am encountering...
     
    Last edited by a moderator: Sep 14, 2015
  2. jcsd
  3. Sep 14, 2015 #2

    DEvens

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  4. Sep 14, 2015 #3
    So I did this: $$ V = \frac {kq} {r}; 9851 Volts = \frac { 8.9 * 10^9 * q } { 0.0458 m } ; Hence, q = 5.069 C $$
    Now since I know the ## q ## I decided to find ## V ## at ## 9.1 r ##.
    To do so I followed these steps: $$ V = \frac { 8.9 * 10^9 * 5.069 } { 9.1 * 0.0458 } = 1.08 * 10^11 V $$
    My question: Why is that the given Potential is smaller than the potential at ## 9.1 * r ## ?
    Shouldn't it be smaller since ## V = \frac { U } { q } ## and ## U = \frac {k q} { r } ## ? :/
     
    Last edited: Sep 14, 2015
  5. Sep 14, 2015 #4

    jtbell

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    Staff: Mentor

    Double-check your arithmetic in this step.
     
  6. Sep 14, 2015 #5
    Apparently, I forgot to put a parenthesis while dividing by 8.9e9. jtbell, thank you!

    So, q = 5.07e-7
    Hence, ## \frac {\Delta V} {\Delta r} = -2.36e4 V/m ##

    I am hoping this answer is correct.
     
  7. Sep 14, 2015 #6

    jtbell

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    Your q is still off by a factor of 10. Check the decimal points in your input numbers. I used the numbers in the first equation of post #3.
     
  8. Sep 14, 2015 #7
    Yes, I see. I wrote -7, instead of -8. The final answer that I have gotten is calculated with 5.069*10^(-8). Thank you!
     
  9. Sep 14, 2015 #8

    jtbell

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    Now for the second part (the gradient), how did you calculate that number, exactly? It's off by (approximately, not exactly) a factor of 10 also.
     
  10. Sep 14, 2015 #9
    Hmm. What's going on lol.

    Okay, let me work on it again.
    so if q = 5.069 e -8

    Now I need to find V at distance 9.1*r (away from center of sphere)
    So V = kq/9.1*r = k*5.069 e -8 / (9.1*0.0458) = 1082.44 Volts

    Potential Gradient = delta V / delta x = (1082.44 V - 9851 V) / (9.1*r - r) | 9.1*r - r = r ( 9.1 - 1) = 8.1*r

    PG = - 23636.196 V/m = -2.36 e 4 V/m

    Am I still wrong?
     
  11. Sep 14, 2015 #10

    jtbell

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    I would call that the "average gradient" between the two r's. It's analogous to average velocity = (x2 - x1) / (t2 - t1).

    The problem asks for the gradient at the second value of r (9.1r). This is analogous to the instantaneous velocity v = dx/dt. As DEvens noted in post #2, "gradient" here means "derivative with respect to r".
     
  12. Sep 14, 2015 #11
    If ## V (at 9.1r) = \frac {kq} {9.1*r} ## then ## \frac {dV} {dr} = \frac {kq} {9.1} *(\frac {-1} {r^2}) ## here ## q = \frac {V_r * r} {k} ## which I plugged in to the equation and got ## \frac {dV} {dr} = \frac { -V_r} {9.1*r} ## which makes ## \frac {dV} {dr} = -23635.97 = -2.36e10^4 ##

    What's wrong..
     
    Last edited: Sep 14, 2015
  13. Sep 14, 2015 #12

    jtbell

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    Take the derivative before substituting 9.1r (0.4168m) for r. This gives you an equation for the gradient at any r (provided that it's ≥ the radius of the sphere). Then substitute your desired value of r.
     
  14. Sep 14, 2015 #13
    I did it before subbing a number for r though? And r is not 0.4168 m it's 0.0458 m ( in the question it says its r = 4.58 cm )
     
  15. Sep 14, 2015 #14

    jtbell

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    When you take the derivative, you need to use an equation with a "generic" r, that can have any value. Your problem statement is confusing because it uses r and 9.1r to refer to "specific" values of r. I would have written it more carefully by calling the radius of the sphere r1 and the other radius r2 = 9.1r1.
     
  16. Sep 14, 2015 #15
    I got -2.15e5.

    Is this what you also got?
     
    Last edited: Sep 14, 2015
  17. Sep 14, 2015 #16

    jtbell

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    That's it. :biggrin:
     
  18. Sep 17, 2015 #17
    How did you guys get this result because I did it over and over with the values provided and I got something else. Here's a summary of the values used according to my calculations:

    ro=0.0458 m
    r=9.1r0= 0.41678 m
    Q=5.02*10-8 C → from Vsurface
    dV/dr=-kq/r2 V/m

    ⇒dV/dr= (8.99*109) N*m2/C2 * (5.02*10-8) C / (0.41678)2 m2

    ⇒dV/dr=-2597.36≅-2.60*103 V/m

     
  19. Sep 22, 2015 #18
    There is a vague step which I did not fully understand at first due to weak math background.

    Use "the definition of derivative" and it will be much clear and you will most likely get a correct result
     
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